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Use a triple integral to find the volume of the solid bounded by the parabolic cylinder enter image description here and the planesenter image description here and enter image description here

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2 Answers 2

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The range of z is obviously from z = 0 to z = 6.

The range of y is obviously from y = 0 to y = 5, with the lower bound being 0 because $2x^2$ us always greater than 0.

The range of x is a little trickier but can be easily determined by solving $y = 2x^2$ and determining the upper and lower expressions for the value of x.

$$ V = \int_0^6\int_0^5\int_{-\sqrt{\frac{y}{2}}}^\sqrt{\frac{y}{2}} dxdydz\, \\ V = \int_0^6\int_0^5\ \sqrt{\frac{y}{2}} - (- \sqrt{\frac{y}{2}}) dydz\,\\ V = \int_0^6\int_0^5\ \sqrt{2}\sqrt{y} dydz\,\\ V = \int_0^6 \sqrt{2}\frac{2}{3}y^\frac{3}{2} \left.\right\vert_{0}^{5}dz\,\\ V = \int_0^6 \sqrt{2}\frac{2}{3}\sqrt{5^3}dz\,\\ V = 6\sqrt{2}\frac{2}{3}\sqrt{5^3}\,\\ V = 20\sqrt{10} \\ $$

Note that the value being integrated (after the bounds have been determined) is 1. This could change if you're integrating over some density function where the weighting of every point with the object will vary.

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  • $\begingroup$ well, now after finding the limits what do we integrate??? $\endgroup$
    – user131040
    Mar 13, 2014 at 3:47
  • $\begingroup$ @user131040 You integrate 1. dxdydz = 1dxdydz $\endgroup$ Mar 13, 2014 at 3:52
  • $\begingroup$ Oh, sorry if that wasn't clear. I'm in the process of adding my work but @zerosofthezeta is correct. $\endgroup$
    – Olshansky
    Mar 13, 2014 at 3:53
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The integral set up for the volume V is:

$$\int_ 0 ^5\int_{-(y/2)^{.5} } ^{(y/2)^{(.5)}}\int_ 0 ^ 6dzdxdy$$. You can

integrate with respect to z first, then x , and then y.

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  • $\begingroup$ With a reputation of 2k+, I wonder why you did not use LaTex? Was it on purpose? $\endgroup$ Mar 13, 2014 at 3:38
  • $\begingroup$ @zerosofthezeta: brutally honestly, I read a few tutorial pages but need a tutor to sit down with me to walk me through it. $\endgroup$
    – DeepSea
    Mar 13, 2014 at 3:40
  • $\begingroup$ I got 2sqrt10 for the final answer which is wrong! $\endgroup$
    – user131040
    Mar 13, 2014 at 3:40
  • $\begingroup$ @user131040: will fix it $\endgroup$
    – DeepSea
    Mar 13, 2014 at 3:46

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